Let f (x) = ax ^ 2 + BX + C (a > 0) and f (1) = - A / 2 (1) prove that f (x) has two zeros

Let f (x) = ax ^ 2 + BX + C (a > 0) and f (1) = - A / 2 (1) prove that f (x) has two zeros

It is proved that: F (1) = a + B + C = - A / 2
∴3a+2b+2c=0．
∴c=-3a /2 -b．
∴f(x)=ax^2+bx-3a /2 -b．
Discriminant △ = B ^ 2-4a (- 3A / 2-B) = B ^ 2 + 6A ^ 2 + 4AB
=（2a+b）^2+2a^2
And ∵ a > 0
Therefore, the function f (x) has two zeros

If the image of the function y = | x-a | + | X-1 | is symmetric with respect to the line x = - 1, then the value of the real number a is______ ．

Because the image shape of the function added by two absolute values is shown in the figure, that is, it is symmetrical with respect to the line where half of the abscissa corresponding to the two turning points lies. And because the image of the function f (x) = | X-1 | + | x-a | = is symmetrical with respect to the line x = - 1, there is a + 12 = - 1, the solution is & nbsp; a = - 3, so the answer is - 3

The real numbers a, B and C are the three numbers in the definition field of y = f (x)
The real numbers a, B and C are the three numbers in the definition field of y = f (x), which satisfy a

f(a)*f(b)

If the real numbers a, B and C are three numbers in the domain of definition of y = f (x) and satisfy a < B < C, f (a) f (b) < 0, f (c) f (b) < 0, then the number of zeros of y = f (x) in the interval (a, c) is ()
A. 2B. Odd C. even D. at least 2

By the existence theorem of roots, f (a) f (b) < 0, f (c) f (b) < 0, then y = f (x) has at least one zero point on the interval (a, b), at least one zero point on (B, c), and f (b) ≠ 0, so the number of zeros of y = f (x) on the interval (a, c) is at least two